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Asked by porayilsanika 11th December 2019, 2:16 PM
If the ball is thrown at angle 30° with horizontal at a speed of 20 m/s, then vertical component of velocity = 20 sin30 = 10 m/s

If the ball which is thrown at a height 2 m hits the wicket which is 0.5 m above ground, then vertical displacement = -1.5 m

we have the formula " S = ut - (1/2)g t2 " ,

where S = vertical displacement = -1.5 m,

u = initial vertical component velocity = 10 m/s

t = time duration to make net vertical displacement -1.5 m
g = acceleration due to gravity

Hence we get,  -1.5 = 10t - (1/2)×9.8×t2   or   4.9 t2 - 10 t - 1.5 = 0  ................(1)

solution of quadratic eqn.(1) is given by :-  t = { 10 + [ 100 + 29.4 ]1/2 } / 9.8 = 2.18 s

Hence time duration for the ball to reach and hit the wicket is 2.18 s.

If Horizontal component of velocity of the ball is ( 10 co30 ) m/s , then horizontal distance d travelled by the

ball is given by,    d  = 10 × cos30 × 2.18 = 18.88 m

Hence fielder is at a distance 18.88 m from wicket

Answered by Expert 11th December 2019, 9:41 PM
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